Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Answer:

Step-by-step explanation:
The relevant rule of exponents is ...
(a^b·c^d)^e = a^(be)·c^(de)
Then ...
(m^(5/4)·n^(-4/5))^(7/3) = m^(5/4·7/3)·n^(-4/5·7/3)
= m^(35/12)·n^(-28/15)
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Since you want positive rational exponents, you can write this as ...
= m^(35/12)/n^(28/15)
Answer:
x=4
Step-by-step explanation:
Remove the parentheses and turn the 4-2x and make it into 4+2x candle the equal term -4 and should be left with x+6=2+2x then move the variable to the left and change the sign so it would be x-2x=2-6 and collect like terms so it would be -x=2-6 and then calculate that and get -x= -4
(sorry its alot)
When X = -9 Y = 1 you can find this by going to -9 on the x axis and finding where the line is there
Use cymath it gives you the answer :)